# Field Theory - The Degree of alpha over F

• Sep 27th 2013, 01:34 AM
Bernhard
Field Theory - The Degree of alpha over F
I am trying to clarify my understanding of Proposition 11 of Dummit and Foote Ch13 Field Theory concerning the degree of $\displaystyle \alpha$ over F.

---------------------------------------------------------------------------------------------------------------------------------------------------------

Proposition 11. Let $\displaystyle \alpha$ be algebraic over the field F and let $\displaystyle F(\alpha)$ be the field generated by $\displaystyle \alpha$ over F.

Then $\displaystyle F(\alpha) \cong F[x]/(m_{\alpha}(x))$

so that in particular $\displaystyle [F(\alpha) \ : \ F] = deg \ m_{\alpha}(x) = deg \ \alpha$

i.e. the degree of $\displaystyle \alpha$ over F is the degree of the extension it generates over F

--------------------------------------------------------------------------------------------------------------------------------------------------------------

However Corollary 7 (Dummit and Foote page 518) states the following:

-----------------------------------------------------------------------------------------------------------------------------------------------------------------

Corollary 7. Suppose in Theorem 6 that p(x) is of degree n. Then

$\displaystyle F(\alpha) = \{ a_0 + a_1 {\alpha} + a_2 {\alpha}^2 + ... ... a_{n-1} {\alpha}^{n-1} \ | \ a_0, a_1, ... ... a_{n-1} \in F \} \subseteq K$

-------------------------------------------------------------------------------------------------------------------------------------------------------------------

Given that $\displaystyle F(\alpha)$ consists of polynomials of degree (n-1) should not the degree of $\displaystyle [F(\alpha) \ : \ F] = deg \ m_{\alpha}(x) - 1 = deg \ \alpha -1$ - that is the degree of $\displaystyle \alpha$ over F be one less than the degree of the minimal polynomial?

Can someone please clarify this situation for me

Peter
• Sep 27th 2013, 03:02 AM
Bernhard
Re: Field Theory - The Degree of alpha over F
I have just been reflecting about my own question above.

It is possible that $\displaystyle F(\alpha)$ viewed as a vector space of deg (n-1) polynomials would have the following as a basis:

$\displaystyle 1, {\alpha}, {\alpha}^2, {\alpha}^3, ... ... {\alpha}^{n-1}$

Then the degree of the space would be n which would equal the degree of the minimal polynomial involved.

If this thought is actually correct, could someone please confirm this as the case.

Peter
• Sep 29th 2013, 05:17 AM
SlipEternal
Re: Field Theory - The Degree of alpha over F
That is correct. Any higher degree polynomials would contain a linear dependency (some linear combination with nonzero coefficients would sum to zero).
• Sep 30th 2013, 07:54 PM
Bernhard
Re: Field Theory - The Degree of alpha over F
Thanks SlipEternal

Sorry my thank you was so late - my day job intervened! :-(